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Step-by-Step Solution
Step 1: Identify the Physical Quantities
We have two circular coils made from the same wire. Let their radii be
$r_1$ and $r_2$, and the currents flowing through them be
$I_1$ and $I_2$, respectively. We also denote their resistances by
$R_1$ and $R_2$, and the potential differences applied across them by
$V_1$ and $V_2$ respectively.
Step 2: Write the Magnetic Field at the Centre of Each Coil
The magnetic field at the centre of a circular coil of radius $r$
carrying current $I$ is given by the formula:
$B = \frac{\mu_0\, I}{2\,r}\,.$
Hence, for Coil 1 and Coil 2, the fields at the centre are:
$B_1 = \frac{\mu_0\, I_1}{2\, r_1}, \quad B_2 = \frac{\mu_0\, I_2}{2\, r_2}\,.$
Step 3: Impose the Condition for Equal Magnetic Fields
Since the magnetic fields at the centres of both coils are the same,
$B_1 = B_2$. Therefore,
$\frac{\mu_0\, I_1}{2\, r_1} = \frac{\mu_0\, I_2}{2\, r_2}\,.$
Given that $r_1 = 2\,r_2$, we substitute this into the above equation:
$\frac{I_1}{2\,r_2} = \frac{I_2}{r_2} \; \Longrightarrow \; I_1 = 2\, I_2\,.$
Step 4: Relate the Resistances of the Coils
The wire for both coils is the same (same material and cross-sectional area), but the length
(and hence the resistance) depends on the radius. Resistance $R$ is proportional to
the total length of the wire, and the circumference is proportional to $r$. Thus:
$R \propto \text{length} \propto r\,.$
Since $r_1 = 2\,r_2$, we get $R_1 = 2\,R_2$.
Step 5: Use Ohm’s Law to Relate the Potential Differences
By Ohm’s Law, $V = IR$. The ratio of the potential differences across the two coils is:
$\frac{V_1}{V_2} = \frac{I_1\, R_1}{I_2\, R_2}\,.$
Substituting $I_1 = 2\,I_2$ and $R_1 = 2\,R_2$:
$\frac{V_1}{V_2} = \frac{(2\,I_2)\,(2\,R_2)}{I_2\, R_2} = 4\,.$
Step 6: State the Final Ratio
The required ratio of potential differences is
$V_1 : V_2 = 4:1$. Therefore, the ratio is 4.
